Large Borel antichains in the Cantor cube?

Question

Let $2^\omega$ be the Cantor cube $\{0,1\}^\omega$, endowed with the standard compact metrizable topology and the standard product measure, called the Haar measure. The Cantor cube is considered as a partially ordered set endowed with the partial order $\le$ defined by ($f\le g$ iff $f(n)\le g(n)$ for all $n\in\omega$).

A subset $A\subset 2^\omega$ is called an antichain if any two distinct elements of $A$ are incomparable in this partial order.

General Problem. How large can a Borel antichain in $2^\omega$ be?

This question can be made more precise using the language of $\sigma$-ideals.

By $\mathcal M$, $\mathcal N$ and $\mathcal E$ we denote the $\sigma$-ideal of meager subsets of $2^\omega$, the $\sigma$-ideal of Haar-null sets in $2^\omega$, and the $\sigma$-ideal generated by closed Haar-null sets, respectively.

It is clear that $\mathcal E\subset\mathcal M\cap\mathcal N$.

Theorem. Each Borel antichain in $2^\omega$ belongs to the $\sigma$-ideal $\mathcal M\cap\mathcal N$.

Proof. Identify $2^\omega$ with the compact topological group $\mathbb Z_2^\omega$ and apply the classical Steinhaus or Piccard-Pettis Theorem, which implies that the difference $A-A$ is a neighborhood of zero and hence contains the characteristic function $\chi$ of some singleton $\{n\}$. Choose any elements $a,b\in A$ with $a-b=\chi$ and conclude that they are comparable in the partial order $\le$ on $2^\omega$.

Now we state more

Precise Problem. Does every Borel antichain $A\subset 2^\omega$ belong to the $\sigma$-ideal $\mathcal E$?

This problem is equivalent to an even

More Precise Question. Assume that a $G_\delta$-set $A$ is an antichain in $2^\omega$. Can its closure $\bar A$ in $2^\omega$ have positive Haar measure?

Answer 1

The answer to the More Precise Question is yes. Define

$$B_n=\{f\in 2^\omega:f(i)\neq f(2^n+i)\text{ for all }0\leq i<2^n\}$$ $$C_n= \{f\in 2^\omega:f(i)= f(2^n+i)=0\text{ for some }0\leq i<2^n\}$$ $$A=\bigcap_{n\geq 10} (B_n\cup C_n) \cap \bigcap_{m\geq 0}\bigcup_{n\geq m}B_n$$

Consider pairs $f,g\in A$ with $f\leq g.$ Let $i\in \omega$ be arbitrary, and pick $n\geq 10$ with $i<2^n$ and such that $f\in B_n.$ Elements of $C_n$ cannot be greater than or equal to elements of $B_n,$ so we must have $g\in B_n.$ We get $1-f(i)=f(2^n+i)\leq g(2^n+i)=1-g(i),$ but also $f(i)\leq g(i),$ so $f(i)=g(i).$ Since $i$ was arbitrary, $f=g.$

So $A$ is a $G_\delta$ antichain. I claim that $A$ is a dense subspace of $X:=\bigcap_{n\geq 10}(B_n\cup C_n).$ A quick argument is that each $\bigcup_{n\geq m} B_n$ is a dense open subset of $X,$ so $A$ is comeagre in $X.$ Alternatively, given an element $f\in X$ and an integer $N\geq 10,$ define recursively $$f_N(i)=\begin{cases}f(i)&(0\leq i<2^N)\\ 1-f_N(i-2^{\lfloor \log_2 i\rfloor})&(i\geq 2^N).\end{cases}$$ The $i<2^N$ case of this definition ensures $f_N\in \bigcap_{n\geq 10}^{N-1}(B_n\cup C_n),$ and the $i\geq 2^N$ case ensures $f_N\in \bigcap_{n\geq N}B_n.$ So each $f_N$ is in $A,$ and $f_N\to f.$ This proves that $\overline{A}=X.$

In particular, $\overline{A}$ contains the subspace $\bigcap_{n\geq 10}C_n.$ This has positive measure because its complement has measure at most $\sum _{n\geq 10}(3/4)^{2^n}<1.$